An object is thrown vertically from a height of 5 "m" at 3 "m/s". How long will it take for the object to hit the ground?
1 Answer
Explanation:
"Thrown vertically" could mean either upwards or downwards. I am assuming it to be downwards.
Step 1: Understand the problem.
An object is under free fall. It is accelerating at
Step 2: Identify the relevant equations.
To find displacement,
s = v_0 t + 1/2 a t^2
To get the time as a function of the distance traveled, make
t^2 + frac{2 v_0}{a}t - frac{2 s}{a} = 0
Subsequently, complete the square.
(t + frac{v_0}{a})^2 - frac{2 s}{a} = (frac{v_0}{a})^2
(t + frac{v_0}{a})^2 = frac{v_0^2}{a^2} + frac{2 s}{a}
= frac{v_0^2 + 2as}{a^2}
Reject the negative square root and take the positive one. This is because we only restrict ourselves to
t + frac{v_0}{a} = frac{sqrt{v_0^2 + 2as}}{a}
t = frac{sqrt{v_0^2 + 2as} - v_0}{a}
Step 3: Set up the coordinate system.
I let the downwards direction be positive. The ground level is arbitrarily set to be the origin (i.e.
Step 4: Identiy the parameters.
In this question,
-
a = g = 9.81 quad "m/s"^2 -
v_0 = 3 quad "m/s" -
s = 5 quad "m"
Step 5: Plug into the equation.
t = frac{sqrt{v_0^2 + 2as} - v_0}{a}
= frac{sqrt{(3 quad "m/s")^2 + 2(9.81 quad "m/s"^2)(5 quad "m")} - (3 quad "m/s")}{9.81 quad "m/s"^2}
= 0.749 quad "s"
Now earlier, I assume that the object has an initial velocity of
v_0 = - 3 quad "m/s"
Plugging in the values again
t = frac{sqrt{v_0^2 + 2as} - v_0}{a}
= frac{sqrt{(-3 quad "m/s")^2 + 2(9.81 quad "m/s"^2)(5 quad "m")} - (-3 quad "m/s")}{9.81 quad "m/s"^2}
= 1.361 quad "s"
